Math Problem Statement

Convert (1) (45113)6 ( )10

Solution

To convert the base-6 number 45113645113_6 to a base-10 (decimal) number, follow these steps:

Step 1: Write the base-6 number in expanded form.

Each digit represents a power of 6, starting from the rightmost digit (which is 606^0).

451136=464+563+162+161+36045113_6 = 4 \cdot 6^4 + 5 \cdot 6^3 + 1 \cdot 6^2 + 1 \cdot 6^1 + 3 \cdot 6^0

Step 2: Calculate the powers of 6.

60=1,61=6,62=36,63=216,64=12966^0 = 1,\quad 6^1 = 6,\quad 6^2 = 36,\quad 6^3 = 216,\quad 6^4 = 1296

Step 3: Multiply the digits by the powers of 6 and sum the results.

451136=(41296)+(5216)+(136)+(16)+(31)45113_6 = (4 \cdot 1296) + (5 \cdot 216) + (1 \cdot 36) + (1 \cdot 6) + (3 \cdot 1) =5184+1080+36+6+3= 5184 + 1080 + 36 + 6 + 3 =6309= 6309

Thus, 45113645113_6 in base-10 is: 451136=63091045113_6 = 6309_{10}

Final Answer:

(45113)6=(6309)10(45113)_6 = (6309)_{10}

Would you like further details or explanations?

Here are 5 related questions to consider:

  1. How would you convert a base-10 number to base-6?
  2. Can you explain why each position in base-6 is a power of 6?
  3. How would this process change for other bases, like base-8 or base-12?
  4. What are some practical applications of base conversions?
  5. How would you convert fractional numbers from base-6 to base-10?

Tip: Always double-check the powers and multiplications when converting between bases to avoid small errors!

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Math Problem Analysis

Mathematical Concepts

Number Systems
Base Conversion

Formulas

n_b = d_0 * b^0 + d_1 * b^1 + ... + d_n * b^n, where n_b is a number in base b and d_i are its digits
Powers of the base for positional values (e.g., 6^0, 6^1, etc.)

Theorems

Positional Notation Theorem

Suitable Grade Level

Grades 7-10